Difference between revisions of "2001 AIME I Problems/Problem 4"
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− | Let <math>D</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AB}</math>. By simple angle-chasing, we find that <math>\angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT</math>, and thus <math>AC = AT = 24</math>. Now <math>\triangle ADC</math> is a <math>30-60-90</math> [[right triangle]] and <math>BDC</math> is a <math>45-45-90</math> right triangle, so <math> | + | Let <math>D</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AB}</math>. By simple angle-chasing, we find that <math>\angle ATB = 105^{\circ}, \angle ATC = 75^{\circ} = \angle ACT</math>, and thus <math>AC = AT = 24</math>. Now <math>\triangle ADC</math> is a <math>30-60-90</math> [[right triangle]] and <math>BDC</math> is a <math>45-45-90</math> right triangle, so <math>AD = 12,\,CD = 12\sqrt{3},\,BD = 12\sqrt{3}</math>. The area of |
<cmath>ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},</cmath> | <cmath>ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},</cmath> | ||
− | and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>. | + | and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>. |
== See also == | == See also == |
Revision as of 08:51, 15 March 2010
Problem
In triangle , angles
and
measure
degrees and
degrees, respectively. The bisector of angle
intersects
at
, and
. The area of triangle
can be written in the form
, where
,
, and
are positive integers, and
is not divisible by the square of any prime. Find
.
Solution
![[asy] size(180); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),B=(12+12*3^.5,0),C=(12,12*3^.5),D=foot(C,A,B),T=IP(CR(A,24),B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(D(MP("T",T,NE))--A); D(MP("D",D)--C,linetype("6 6") + linewidth(0.7)); MP("24",(A+3*T)/4,SE); D(anglemark(C,B,A,65)); D(anglemark(B,A,C,65)); D(rightanglemark(C,D,B,50)); MP("30^{\circ}",A,(4,1)); MP("45^{\circ}",B,(-3,1)); [/asy]](http://latex.artofproblemsolving.com/f/d/8/fd80ef0be57cda8d5c014431604ac07fc2c9a044.png)
Let be the foot of the altitude from
to
. By simple angle-chasing, we find that
, and thus
. Now
is a
right triangle and
is a
right triangle, so
. The area of
and the answer is .
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |